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adell [148]
3 years ago
6

You Have 2 2/3 Cups Of Dried Fruit To Divide Evenly Amoung 3 Children. How Many Cups Of Fruit Will Each Child Receive?

Mathematics
1 answer:
kogti [31]3 years ago
6 0
The dried fruit divided evenly between the 3 children, each would get 9/8 cups of dried fruit
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Fractions are hard... Or I'm just to lazy to try and do them.
notka56 [123]
Answer: C) 3/2

Explanation: In math, the reciprocal is the inverse of a number or value. Simply, the number when flipped. Knowing this, one can flip the fraction to find the reciprocal, in this case turning 2/3 into 3/2.

Hope this helps :)
5 0
3 years ago
I need all of these please help
Korvikt [17]
What grade you in and what unit lesson is this, I might be able to help you. I go to the same school.
7 0
3 years ago
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FOR EASY BRAINLIEST ANSWER AND EXPLAIN!! ASAP
Alina [70]
Answer: 9 1/3 minutes
First create an equation with the given information; with x being the commercial time, 30=3x+2, now solve for x. First subtract 2 from each side and get 28=3x and then divide each side by 3 to get 9 and 1/3 minutes.
3 0
3 years ago
How do you write the number 3 2/4 as a fraction greater than 1
tatyana61 [14]
14/4 > 1 is what it looks like.
8 0
3 years ago
In a restaurant, the proportion of people who order coffee with their dinner is p. A simple random sample of 144 patrons of the
mote1985 [20]
<h2>Answer with explanation:</h2>

Given : In a restaurant, the proportion of people who order coffee with their dinner is p.

Sample size : n= 144

x= 120

\hat{p}=\dfrac{x}{n}=\dfrac{120}{144}=0.83333333\approx0.8333

The null and the alternative hypotheses if you want to test if p is greater than or equal to 0.85 will be :-

Null hypothesis : H_0: p\geq0.85   [ it takes equality (=, ≤, ≥) ]

Alternative hypothesis : H_1: p  [its exactly opposite of null hypothesis]

∵Alternative hypothesis is left tailed, so the test is a left tailed test.

Test statistic : z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}

z=\dfrac{0.83-0.85}{\sqrt{\dfrac{0.85(1-0.85)}{144}}}\\\\=-0.561232257678\approx-0.56

Using z-vale table ,

Critical value for 0.05 significance ( left-tailed test)=-1.645

Since the calculated value of test statistic is greater than the critical value , so we failed to reject the null hypothesis.

Conclusion : We have enough evidence to support the claim that p is greater than or equal to 0.85.

8 0
3 years ago
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